Question

我开发了通过pyswip连接到Prolog的python应用程序。

以下是我执行的Prolog部分的示例。

init( Board) :-
  Board = b(n,x,n,x,n,x,n,x,x,n,x,n,x,n,x,n,n,x,n,x,n,x,n,x,e,n,e,n,e,n,e,n,n,e,n,e,n,e,n,e,o,n,o,n,o,n,o,n,n,o,n,o,n,o,n,o,o,n,o,n,o,n,o,n).

执行以下代码时

soln = list(self.prolog.query("init(B)"))
print(soln[0])

soln [0]是

<'B': 'Functor6905740'>

如何获得b（n，x，n，x，...）而不是Functor参考？

聚苯乙烯。我使用所有64位：Python 2.7，SWI-Prolog，pyswip，Visual Studio 2013

Answer 1

我找到了自己的解决方案我导入Functor，Variable，Query

from pyswip import Prolog, Functor, Variable, Query

要获得仿函数值，我使用

init = Functor("init", 1)
B = Variable()
soln = Query(init(B))

while soln.nextSolution():
   b64 = B.get_value()
self.Board_pl = "b("
for i in range (0, 64):
    self.Board_pl += (str(b64.args[i]) + ",")
self.Board_pl += (str(b64.args[63]) + ")")
print(Board)

然后，程序将打印

b(n,x,n,x,n,x,n,x,x,n,x,n,x,n,x,n,n,x,n,x,n,x,n,x,e,n,e,n,e,n,e,n,n,e,n,e,n,e,n,e,o,n,o,n,o,n,o,n,n,o,n,o,n,o,n,o,o,n,o,n,o,n,o,n)

如何使用pyswip在Prolog中获取Functor字符串

1 个答案: